We have no idea what the actual cost is. All we know is that it’s at least 20, but since h(n)
can be an underestimate and g(n) is perfect, f(n) can also be an underestimate
5.
1 point each
Are these heuristics admissible (T/F)? Note: C(n, goal) is the true cost from the node to the goal.
a.
h1(n)=0
T
b.
h2(n)=g(n)
F
c.
h3(n)= random(0, C(n, goal))
T
d.
h4(n)=2C(n, goal)
F
e.
h5(n)=½C(n, goal)
T
Most people got this with the exception of 5b, g(n). As you get closer to the solution, g
grows and h shrinks. If you used g(n) as
your heuristic, h(n) gets bigger as
you get closer
to the goal, which is the wrong direction. Suppose you’re on a path C*=20. If you’re at the
start of the path, you might have g(n)=5, h(n)=15. Near the end of the path, g(n)=15,
h(n)=5. If you used h(n)=g(n) then near the end of the path the remaining cost is 5you’re
your estimate is 15, 3 times the actual cost and clearly not admissible. And what happens
with that inadmissible heuristic? You’d have f(n)=g(n)+h(n) = 15+15 = 30 which is > C*, so
the solution would be below where it should be and a non-optimal solution might be found
first. It also has the effect of pushing A* away from the goal, not towards it
6.
4 points
The open list contains 101 nodes. One is the goal, f(goal)=45. Of the remaining 100 nodes, 10
have the cost f(n)=10, 10 have f(n)=20, 10 have f(n)=30, etc. all the way to 10 nodes with
f(n)=100. Of these, only 8 nodes are on the optimal path. Including the goal node, how many of